Question

Two chords $AB$ and $CD$ of a circle intersect each other at $P$ outside the circle. If $AB=6cm,BP=2cm$ and $PD=2.5cm$, find $CD$.

Answer 1

It is given that $AB$ and $CD$ of a circle intersect each other at $P$ outside the circle. If $AB=6cm,BP=2cm$ and $PD=2.5cm$

So we get

$AP\times BP=CP\times DP$

From the figure we know that $CP=CD+DP$

By substituting the values

$8\times 2=(CD+2.5)\times 2.5cm$

consider $x=CD$

So we get

$8\times 2=(x+2.5)\times 2.5$

On further calculation

$16=2.5x+6.25$

It can be written as

$2.5x=16-6.25$

By subtraction

$2.5x=9.75$

By division

$x=\frac{9.75}{2.5}$

So we get

$x=3.9cm$

Therefore, $CD=3.9cm$.